Problem: $\text B = \left[\begin{array}{rr}1 & 4 \\ 1 & 3\end{array}\right]$ and $\text D = \left[\begin{array}{rr}4 & 3 \\ 0 & -1\end{array}\right]$. Let $\text {H = BD}$. Find $\text H$. $ {H = }$
Explanation: The Strategy When multiplying matrices, we should find each entry of the resulting product matrix separately. To find entry $(i,j)$ of the resulting product matrix, we calculate the vector dot product of row $i$ of the first matrix and column $j$ of the second matrix. [I don't know what "vector dot product" is!] Finding $\text {H}_{1,1}$ $\text{H}_{1,1}$ is the dot product of the first row of $\text{B}$ and the first column of $\text{D}$. $ \text {H}=\left[\begin{array}{rr}{1} & {4} \\ 1 & 3\end{array}\right]\left[\begin{array}{rr} {4} & 3 \\ {0} & -1\end{array}\right]$ Therefore, this is the appropriate calculation of $\text{H}_{1,1}$. $\begin{aligned}\text{H}_{1,1}&=(1,4)\cdot(4,0)\\\\ &=1 \cdot 4 + 4\cdot 0\\\\ &=4 \end{aligned}$ The other entries of $\text{H}$ can be found similarly. Try it yourself for $\text{H}_{2,1}$ What is the appropriate calculation of ${H}_{2,1}$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $4 \cdot 4 + 3\cdot 3= 25$ (Choice B) B $1 \cdot 4 + 3\cdot 0 = 4$ (Choice C) C $1 \cdot 3 + 4\cdot -1 = -1$ Check Summary After calculating all the remaining entries of $\text{H}$, we get the following answer. $ \text{H}=\left[\begin{array}{rr}4 & -1 \\ 4 & 0\end{array}\right]$